Intereting Posts

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Let $G$ be a finite abelian group, $G = \{e, a_{1}, a_{2}, …, a_{n} \}$. Prove that $(a_{1}a_{2}\cdot \cdot \cdot a_{n})^{2} = e$.

I’ve been stuck on this problem for quite some time. Could someone give me a hint?

Thanks in advance.

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Here is a hint: for any given $a_i\in G$, there are two possibilities: either

- $a_i$ is its own inverse, or
- $a_i$ is not its own inverse, but rather $a_j=a_i^{-1}$ for some $j\neq i$.

The map $\phi:x\in G\mapsto x^{-1}\in G$ is an automorphism of $G$ so, in particular, it induces a bijection $G\setminus\{e\}\to G\setminus\{e\}$. It maps $b=a_1\cdots a_n$ to itself, so that $b=b^{-1}$ and, therefore, $b^2=e$.

Note that a question which is in some sense more natural is: “What is the product of all the elements of a finite abelian group?” The given question gives some information about this product, namely that it is an element of order at most $2$.

The answer to the more general question is given by the following result.

**Wilson’s Theorem in a Finite Abelian Group**: Let $G$ be a finite abelian group. Then the product of all elements of $G$ is equal to the identity *unless* $G$ has exactly one element $t$ of order $2$, in which case the product is $t$. For a proof see e.g. $\S 6$ of these notes.

In fact one can view the answer to the OP’s question as 2/3 of the way towards a proof of WTFAG. Namely, we take the product of all of the elements in the group, and note that the elements which have order greater than $2$ occur in pairs $x,x^{-1}$. Therefore the product of all the elements in a finite abelian group $G$ is also equal to the product of all the elements of $G[2]$, i.e., the product of all elements of order at most $2$. In an abelian group, the subset $G[2]$ is a subgroup, so the product has order $2$ and hence squares to the identity element $e$, completing the answer to the OP’s question.

But now, on to WTFAG: First of all, if there are no elements of order $2$ then $G[2] = \{e\}$, verifying WTFAG in that case. Second of all, if there is exactly one element $t$ of order $2$, then $G[2] = \{e,t\}$ and the product of all elements in $G[2]$ is equal to $t$.

To complete the proof of WTFAG, one needs to consider the case in which $G[2]$ has more than two elements. In my notes, I do this by first establishing the CLAIM that $G[2]$ (or really any finite group in which every element is self-inverse) is isomorphic to a finite direct product of copies of groups of order $2$, and then using some simple counting arguments.

The CLAIM follows either from the structure theory for finite abelian groups — which is proved in the same set of notes, but nevertheless the emphasis in the notes is on the fact that in many applications, especially in number theory, the big structure theorem can be avoided — or from the fact that a group in which each element has order $2$ is necessarily a vector space over the field $\mathbb{F}_2$ of two elements (and then we invoke the structure theorem for finite-dimensional vector spaces). I would actually prefer to have a third, more elementary proof of this fact. Perhaps someone can suggest one?

In addition to Pete Clark’s answer, there is also a very neat answer to the question *What is the set of all different products of all the elements of a finite group $G$?* So $G$ not necessarily *abelian*. Well, if a 2-Sylow subgroup of $G$ is trivial or non-cyclic, then this set equals the commutator subgroup $G'$. If a 2-Sylow subgroup of $G$ is cyclic, then this set is the coset $xG'$ of the commutator subgroup, with $x$ the unique involution of a 2-Sylow subgroup.

HINT:

An element and its inverse are unique, and each element is either its own inverse or not.

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